b) \(\left(\frac{2}{3}x-1\right).\left(\frac{3}{4}x+\frac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x-1=0\\\frac{3}{4}x+\frac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{2}{3}x=1\\\frac{3}{4}x=-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1:\frac{2}{3}\\x=\left(-\frac{1}{2}\right):\frac{3}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=-\frac{2}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}.\)

c) \(x:\frac{9}{14}=\frac{7}{3}:x\)

\(\Rightarrow\frac{x}{\frac{19}{4}}=\frac{\frac{7}{3}}{x}\)

\(\Rightarrow x.x=\frac{7}{3}.\frac{19}{4}\)

\(\Rightarrow x.x=\frac{133}{12}\)

\(\Rightarrow x^2=\frac{133}{12}\)

\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\frac{133}{12}}\\x=-\sqrt{\frac{133}{12}}\end{matrix}\right.\)

Vậy \(x\in\left\{\sqrt{\frac{133}{12}};-\sqrt{\frac{133}{12}}\right\}.\)

d) \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\)

\(\Rightarrow\left(3x-1\right)^{10}-\left(3x-1\right)^{20}=0\)

\(\Rightarrow\left(3x-1\right)^{10}.\left[1-\left(3x-1\right)^{10}\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^{10}=0\\1-\left(3x-1\right)^{10}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-1=0\\\left(3x-1\right)^{10}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x-1=\pm1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1:3\\3x-1=1\\3x-1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\3x=2\\3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{2}{3}\\x=0\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{1}{3};\frac{2}{3};0\right\}.\)

Chúc bạn học tốt!

Mơn bạn nha

a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

b) \(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)

\(\Rightarrow\frac{297}{164}=x:\frac{75}{90}\)

\(\Rightarrow x=\frac{297}{164}.\frac{75}{90}\)

\(\Rightarrow x=\frac{495}{328}\)

Vậy \(x=\frac{495}{328}.\)

c) \(x+\left|-\frac{1}{2}\right|=3\frac{1}{3}-4\frac{1}{2}\)

\(\Rightarrow x+\frac{1}{2}=\frac{10}{3}-\frac{9}{2}\)

\(\Rightarrow x+\frac{1}{2}=-\frac{7}{6}\)

\(\Rightarrow x=\left(-\frac{7}{6}\right)-\frac{1}{2}\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

ths bạn

Cô chào em, chúc em 1 buổi tối vui vẻ bên gia đình! ^^

hòi cô phải giáo viên ko vậy ạ

hi bn

"Thế giới ảo có hàng triệu người, sao bạn lại vinh dự nhận được lời chào từ mình nhỉ?"

chào bạn nha

a) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{a}{b}+1=\frac{c}{d}+1\)

\(\Rightarrow\frac{a}{b}+\frac{b}{b}=\frac{c}{d}+\frac{d}{d}\)

\(\Rightarrow\frac{a+b}{b}=\frac{c+d}{d}\left(đpcm1\right).\)

b) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)

\(\Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}\)

\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\left(đpcm2\right).\)

c) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)

\(\Rightarrow\frac{b}{a}+1=\frac{d}{c}+1\)

\(\Rightarrow\frac{b}{a}+\frac{a}{a}=\frac{d}{c}+\frac{c}{c}\)

\(\Rightarrow\frac{b+a}{a}=\frac{d+c}{c}\left(đpcm3\right).\)

d) Ta có: \(\frac{a}{b}=\frac{c}{d}.\)

\(\Rightarrow\frac{b}{a}=\frac{d}{c}.\)

\(\Rightarrow\frac{b}{a}-1=\frac{d}{c}-1\)

\(\Rightarrow\frac{b}{a}-\frac{a}{a}=\frac{d}{c}-\frac{c}{c}\)

\(\Rightarrow\frac{b-a}{a}=\frac{d-c}{c}\left(đpcm4\right).\)

Còn 2 câu kia tí nữa mình làm sau nhé.

Chúc bạn học tốt!

ths banj

1.x<-1 hoặc x=1

2.x>2

3.x=0 hoặc x=1/7

a/ Giả sử \(\sqrt{3}\) là số hữu tí

\(\Leftrightarrow\sqrt{3}=\dfrac{a}{b}\left(a,b\in N\right)\)

\(\Leftrightarrow\left(\sqrt{3}\right)^2=\left(\dfrac{a}{b}\right)^2\)

\(\Leftrightarrow3=\dfrac{a^2}{b^2}\)

\(\Leftrightarrow a^2=3b^2\)

\(\Leftrightarrow a^2⋮3\) Mà 3 là số nguyên tố

\(\Leftrightarrow a⋮3\)

\(\Leftrightarrow a^2⋮9\)

\(\Leftrightarrow3b^2⋮9\)

\(\Leftrightarrow b^2⋮3\)

\(\Leftrightarrow b⋮3\)

\(\Leftrightarrow\left(a,b\right)\ne1\) \(\Leftrightarrow\) Giả sử sai \(\Leftrightarrow\sqrt{3}\) là số vô tỉ

a) \(\frac{7-8x}{6}=\frac{-4+2x}{5}\)

=> \(\left(7-8x\right).5=6\left(-4+2x\right)\)

=> 35 - 40x = -24 + 12x

=> 35 + 24 = 12x + 40x

=> 52x = 59

=> x = 59/52

b) \(\frac{1-3:x}{8}=\frac{8}{1-3:x}\)

=> (1 - 3: x)² = 8²

=> \(\orbr{\begin{cases}1-3:x=8\\1-3:x=-8\end{cases}}\)

=> \(\orbr{\begin{cases}3:x=-7\\3:x=9\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{7}\\x=\frac{1}{3}\end{cases}}\)

c) \(\left(x+1\right)\left(x-2\right)\ge0\)

=> \(\hept{\begin{cases}x+1\ge0\\x-2\ge0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-2\le0\end{cases}}\)

=> \(\hept{\begin{cases}x\ge-1\\x\ge2\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\le2\end{cases}}\)

=> \(-1\le x\le2\)

h) \(\left(x+1\right)\left(x-3\right)\le0\)

=> \(\hept{\begin{cases}x+1\ge0\\x-3\le0\end{cases}}\)hoặc \(\hept{\begin{cases}x+1\le0\\x-3\ge0\end{cases}}\)

=> \(\hept{\begin{cases}x\ge-1\\x\le3\end{cases}}\) hoặc \(\hept{\begin{cases}x\le-1\\x\ge3\end{cases}}\) (loại)

= \(-1\le x\le3\)

Lời giải:
\((\frac{-3}{2})^2+|-\frac{5}{6}|-1\frac{1}{2}:6+(-\frac{1}{2019})\)

\(=\frac{9}{4}+\frac{5}{6}-\frac{1}{4}-\frac{1}{2019}=\frac{8}{4}+\frac{5}{6}-\frac{1}{2019}=2\frac{1121}{1346}\)

c.ơn ạk